Final answer:
2.90 grams of water remain after the reaction between calcium oxide and water is complete. This is determined by stoichiometry and the excess water calculation based on a 1:1 molar ratio between CaO and H2O.
Step-by-step explanation:
To determine how many grams of water remain after the reaction between calcium oxide and water is complete, we must look at the stoichiometry of the reaction. Calcium oxide reacts with water in a 1:1 molar ratio to produce calcium hydroxide:
CaO(s) + H₂O(l) → Ca(OH)₂(s)
The molar mass of CaO is approximately 56.08 g/mol, and the molar mass of H₂O is approximately 18.02 g/mol. For a 4.50 g sample of CaO, the moles of CaO are:
(4.50 g) / (56.08 g/mol) = 0.0802 mol
Similarly, for a 4.34 g sample of H₂O, the moles of H₂O are:
(4.34 g) / (18.02 g/mol) = 0.2408 mol
Since the reaction requires a 1:1 ratio and we have more moles of H₂O than needed, some water will remain unreacted. The amount of water needed to react completely with 0.0802 mol of CaO is:
(0.0802 mol) × (18.02 g/mol) = 1.44 g of H₂O
Therefore, the amount of water remaining after the reaction is complete is:
(4.34 g initial H₂O) - (1.44 g reacted H₂O) = 2.90 g of H₂O
Thus, 2.90 grams of water remain after the reaction between calcium oxide and water is complete.